Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $n = \dfrac{-4a^2 + 28a + 32}{2a^2 - 72} \times \dfrac{-2a - 12}{4a - 32} $
Solution: First factor out any common factors. $n = \dfrac{-4(a^2 - 7a - 8)}{2(a^2 - 36)} \times \dfrac{-2(a + 6)}{4(a - 8)} $ Then factor the quadratic expressions. $n = \dfrac {-4(a - 8)(a + 1)} {2(a + 6)(a - 6)} \times \dfrac {-2(a + 6)} {4(a - 8)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac { -4(a - 8)(a + 1) \times -2(a + 6)} { 2(a + 6)(a - 6) \times 4(a - 8)} $ $n = \dfrac {8(a - 8)(a + 1)(a + 6)} {8(a + 6)(a - 6)(a - 8)} $ Notice that $(a + 6)$ and $(a - 8)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {8(a - 8)(a + 1)\cancel{(a + 6)}} {8\cancel{(a + 6)}(a - 6)(a - 8)} $ We are dividing by $a + 6$ , so $a + 6 \neq 0$ Therefore, $a \neq -6$ $n = \dfrac {8\cancel{(a - 8)}(a + 1)\cancel{(a + 6)}} {8\cancel{(a + 6)}(a - 6)\cancel{(a - 8)}} $ We are dividing by $a - 8$ , so $a - 8 \neq 0$ Therefore, $a \neq 8$ $n = \dfrac {8(a + 1)} {8(a - 6)} $ $ n = \dfrac{a + 1}{a - 6}; a \neq -6; a \neq 8 $